$解: (2) 如圖②，過點(diǎn)A作AD\perp BC于點(diǎn)D.\because AB = AC，\angle BAC = 120^{\circ}，\therefore AD為邊BC上的中線，\angle ABC=\frac{1}{2}(180^{\circ}-\angle BAC)=30^{\circ}. 在Rt\triangle ABD中，\because AB = 4，\angle ABC = 30^{\circ}，易得AD=\frac{1}{2}AB = 2，\therefore BD^{2}=AB^{2}-AD^{2}=4^{2}-2^{2}=12，\therefore AB\triangle AC=AD^{2}-BD^{2}=2^{2}-12=4 - 12=-8. 過點(diǎn)B作BE\perp AC，交CA的延長線于點(diǎn)E，取AC的中點(diǎn)F，連接BF，\therefore AF=\frac{1}{2}AC = 2，\angle BEA = 90^{\circ}，\therefore \angle ABE=\angle BAC-\angle BEA = 120^{\circ}-90^{\circ}=30^{\circ}. 在Rt\triangle ABE中，\because AB = 4，\angle ABE = 30^{\circ}，易得AE=\frac{1}{2}AB = 2，\therefore BE^{2}=AB^{2}-AE^{2}=4^{2}-2^{2}=12. 在Rt\triangle BEF中，\because BE^{2}=12，EF = AE + AF=2 + 2 = 4，\therefore BF^{2}=BE^{2}+EF^{2}=12 + 16 = 28，\therefore BA\triangle BC=BF^{2}-AF^{2}=28 - 4 = 24.$
$ (3) 如圖③，取AN的中點(diǎn)M，連接BM.\because ON=\frac{1}{3}AO，\therefore設(shè)AM = MN = NO = x.\because AB\triangle AC=AO^{2}-BO^{2}=9x^{2}-BO^{2}=14，\therefore BO^{2}=9x^{2}-14.\because BN\triangle BA=BM^{2}-AM^{2}=BM^{2}-x^{2}=10，\therefore BM^{2}=x^{2}+10.\because BO^{2}+OM^{2}=BM^{2}，\therefore (9x^{2}-14)+4x^{2}=x^{2}+10，$
$ \begin{aligned} 9x^{2}-14 + 4x^{2}&=x^{2}+10 \\ 12x^{2}&=24 \\ x^{2}&=2 \\ \end{aligned}$
$ \therefore BO^{2}=9x^{2}-14=4，\therefore BO = 2，BC = 2BO = 4.$
